11. A roller 1.5 m long has a diameter of 70 cm. How many revolution will it make to level a
playground measuring 50 m x 33 m.
12. The diameter of a roller is 42 cm and its length is 120 cm. If takes 500 complete revolution to
move once to land a playground, find the area of the playground in m².
Answers
Answer:
11.length of roller or height = 1.5 m diameter of roller =0.7 m radius of roller = 0.7/2 m
outer area of cylinder = 2nrh
= 2 x 22/7 x 0.7/2 x 1.5
= 3.3 m sq
area of field = 50x33 = 1650 m sq
no. of revolution taken = 1650/3.3 = 500 revolutions
12.We have the diameter of a cyclindrial roller = 42 cm
⇒ The radius of cyclindrical roller (r) = 42 / 2 = 21 cm
Length of a cyclindrical roller (h) = 120 cm
Curved surface of the roller = 2πrh = 2 × 22 / 7 × 21 × 120 = 15840 cm2
= 15840 / 10000 m2 = 1.584 m2
Area covered by the roller in one complete revolution .
= curved surface of the roller = 1.584 m2
∴ Area of the playground = Area covered by the roller in 500 complete revolutions
= 500 × 1.584 = 792 m2
Step-by-step explanation:
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11 number answer is above in the attachment
12) Radius of the roller (r)=84/2cm=42cm
length of the roller (h)=120cm
∴ Area of the playground levelled in taking 1 complete revolution 2πrh
=2×227×42×120=31680cm^2
∴ Area of the playground
=31680×500=15840000cm^2
15840000/100×100 m^2=1584m^2
